Calculus of Variations Summer Term 2014 - UNI-SB.DE · Calculus of Variations Summer Term 2014 Lecture 12 ... 1 Numerical solution of the Euler-Lagrange equations ... Calculus of

Calculus of VariationsSummer Term 2014

Lecture 12

26. Juni 2014

c© Daria Apushkinskaya 2014 () Calculus of variations lecture 12 26. Juni 2014 1 / 25

Purpose of Lesson

Purpose of Lesson:

To discuss numerical solutions of the variational problems

To introduce Euler’s Finite Difference Method and Ritz’s Method.


Numerical Solutions

§9. Numerical Solutions


Numerical Solutions

Numerical Solutions:

The Euler-Lagrange equations may be hard to solve.

Natural response is to find numerical methods.

1 Numerical solution of the Euler-Lagrange equations

We won’t consider these here (see other courses)

2 Euler’s finite difference method

3 Ritz (Rayleigh-Ritz)

In 2D: Kantorovich’s method


Numerical Solutions Euler’s Finite Difference Method

Euler’s Finite Difference Method

We can approximate our function (and hence the integral) onto afinite grid.

In this case, the problem reduces to a standard multivariablemaximization (or minimization) problem, and we find the solutionby setting the derivatives to zero.

In the limit as the grid gets finer, this approximates theEuler-Lagrange equations.


Numerical Solutions Numerical Approximation

Numerical approximation of integrals:

use an arbitrary set of mesh points

a = x0 < x1 < x2 < · · · < xn = b.

approximate

y ′(xi) =yi+1 − yi

xi+1 − xi=

∆yi

∆xi

rectangle rule

J[y ] =

b∫a

F (x , y , y ′)dx 'n−1∑i=0

F(

xi , yi ,∆yi

∆xi

)∆xi = J[y]

J[·] is a function of the vector y = (y1, y2, . . . , yn).


Numerical Solutions Finite Difference Method (FDM)

Treat this as a maximization of a function of n variables, so that werequire

∂J∂yi

= 0

for all i = 1,2, . . . ,n.

Typically use uniform grid so

∆xi = ∆x =b − a

n.


Numerical Solutions Simple Example

Example 12.1Find extremals for

J[y ] =

1∫0

[12

y ′2 +12

y2 − y]

dx

with y(0) = 0 and y(1) = 0.

The Euler-Lagrange equation y ′′ − y = −1.


Numerical Solutions Simple Example: direct solution

Example 12.1 (direct solution)

E-L equation: y ′′ − y = −1

Solution to homogeneous equation y ′′ − y = 0 is given by eλx

giving characteristic equation

λ2 − 1 = 0,

so λ = ±1

Particular solution y = 1.

Final solution isy(x) = Aex + Be−x + 1 .


Numerical Solutions Simple Example: direct solution

Example 12.1 (direct solution)

The boundary conditions y(0) = y(1) = 0 constrain

A + B = −1

Ae + Be−1 = −1

so A =1− ee2 − 1

and B =e − e2

e2 − 1.

Then the exact solution to the extremal problem is

y(x) =1− ee2 − 1

ex +e − e2

e2 − 1e−x + 1 .


Numerical Solutions Simple Example: Euler’s FDM

Example 12.1 (Euler’s FDM)Find extremals for

J[y ] =

1∫0

[12

y ′2 +12

y2 − y]

dx

Euler’s FDM:

Take the grid xi = i/n, for i = 0,1, . . . ,n so

end points y0 = 0 and yn = 0

∆x = 1/n and ∆yi = yi+1 − yi .

So

y ′i = ∆yi/∆x = n(yi+1 − yi )

andy ′2

i = n2 (y2i − 2yiyi+1 + y2

i+1).



Example 12.1 (Euler’s FDM)Find extremals for

J[y ] =

1∫0

[12

y ′2 +12

y2 − y]

dx

Its FDM approximation is

J[y] =n−1∑i=0

F (xi , yi , y ′i )dx

=n−1∑i=0

12

n2(

y2i − 2yiyi+1 + y2

i+1

)∆x +

(y2

i /2− yi

)∆x

=n−1∑i=0

12

n(

y2i − 2yiyi+1 + y2

i+1

)+

y2i /2− yi

n.


Numerical Solutions Simple Example: end-conditions

Example 12.1 (end-conditions)

We know the end conditions y(0) = y(1) = 0, which imply that

y0 = yn = 0.

Include them into the objective using Lagrange multipliers

H[y] =n−1∑i=0

12

n(

y2i − 2yiyi+1 + y2

i+1

)+

y2i /2− yi

n+ λ0y0 + λnyn.



Example 12.1 (Euler’s FDM)Taking derivatives, note that yi only appears in two terms of theFDM approximation

H[y] =n−1∑i=0

12

n(

y2i − 2yiyi+1 + y2

i+1

)+

y2i /2− yi

n+ λ0y0 + λnyn

∂H[y]

∂yi=

n(y0 − y1) +

y0 − 1n

+ λ0 for i = 0

n(2yi − yi+1 − yi−1) + yi/n − 1/n for i = 1, . . . ,n − 1n(yn − yn−1) + λn for i = n

We need to set the derivatives to all be zero, so we now haven = 3 linear equations, including y0 = yn = 0, and n + 3 variablesincluding the two Lagrange multipliers.

We can solve this system numerically using, e.g., Maple.



Example 12.1 (Euler’s FDM)Example: n = 4, solve

Az = b

where

A =

1.004.25 −4.00 1−4.00 8.25 −4.00

−4.00 8.25 −4.00−4.00 8.25 −4.00

−4.00 4.00 1.001.00

and

b = (0.00, 0.25, 0.25, 0.25, 0.25, 0.00, 0.00)T



Example 12.1 (Euler’s FDM)First n + 1 terms of z give y

Last two terms of z give the Lagrange multipliers λ0 and λn.

Solving the system we get for n = 4

y1 = y3 = 0.08492201040, y2 = 0.1126516464


Numerical Solutions Simple Example: results

Example 12.1 (results)


Numerical Solutions Simple Example: results

Example 12.1 (results)


Numerical Solutions Convergence of Euler’s FDM

Convergence of Euler’s FDM

J[y] =n−1∑i=0

F(

xi , yi ,∆yi

∆xi

)∆x and ∆yi = yi+1 − yi

Only two terms in the sum involve yi , so

∂J∂yi

=∂

∂yiF(

xi−1, yi−1,∆yi−1

∆x

)+

∂

∂yiF(

xi , yi ,∆yi

∆x

)=

1∆x

∂F∂y ′i

(xi−1, yi−1,

∆yi−1

∆x

)+∂F∂yi

(xi , yi ,

∆yi

∆x

)− 1

∆x∂F∂y ′i

(xi , yi ,

∆yi

∆x

)

=∂F∂yi

(xi , yi , y ′i )−∂F∂y ′i

(xi , yi ,

∆yi∆x

)− ∂F

∂y ′i

(xi−1, yi−1,

∆yi−1∆x

)∆x


Numerical Solutions Convergence of Euler’s FDM

Convergence of Euler’s FDM

∂J∂yi

=∂F∂yi

(xi , yi , y ′i )−∂F∂y ′i

(xi , yi ,

∆yi∆x

)− ∂F

∂y ′i

(xi−1, yi−1,

∆yi−1∆x

)∆x

= 0.

In limit n→∞, then ∆x → 0, and so we get

∂F∂y− d

dx

(∂F∂y ′

)= 0

which are the Euler-Lagrange equations.

i.e., the finite difference solution converges to the solution of theEuler-Lagrange equations.


Numerical Solutions Comments

RemarksThere are lots of ways to improve Euler’s FDM

use a better method of numerical quadrature (integration)

trapezoidal rule

Simpson’s rule

Romberg’s method

use a non-uniform grid

make it finer where there is more variation

We can use a different approach that can be even better.


Numerical Solutions Ritz’s Method

Ritz’s Method

In Ritz’s method (called Kantorovich’s method where there is morethan one independent variable), we approximate our functions (theextremal in particular) using a family of simple functions.

Again we can reduce the problem into a standard multivariablemaximization problem, but now we seek coefficients for ourapproximation.



Assume we can approximate y(x) by

y(x) = φ0(x) + c1φ1(x) + c2φ2(x) + · · ·+ cnφn(x)

where we choose a convenient set of functions φj(x) and find thevalues of cj which produce an extremal.

For fixed end-points problem:

Choose φ0(x) to satisfy the end conditions.

Then φj(x0) = φj(x1) = 0 for j = 1,2, . . . ,n

The φ can be chosen from standard sets of functions, e.g. powerseries, trigonometric functions, Bessel’s functions, etc. (but must belinearly independent).



Select{φj}n

j=0

Approximate

yn(x) = φ0(x) + c1φ1(x) + c2φ2(x) + · · ·+ cnφn(x)

Approximate J[y ] ' J[yn] =x1∫x0

F (x , yn, y ′n)dx .

Integrate to get J[yn] = Jn(c1, c2, . . . , cn).

Jn is a known function of n variables, so we can maximize (orminimize) it as usual by

∂Jn

∂ci= 0

for all i = 1,2, . . . ,n.


Numerical Solutions Upper Bounds

Assume the extremal of interest is a minimum, then for theextremal

J[y ] < J[y ]

for all y within the neighborhood of y .

Assume our approximating function yn is close enough to be inthat neighborhood, then

J[y ] 6 J[yn] = Jn[c]

so the approximation provides an upper bound on the minimumJ[y ].

Another way to think about it is that we optimize on a smaller setof possible functions y , so we can’t get quite as good a minimum.


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Calculus of Variations Summer Term 2014 - UNI-SB.DE · Calculus of Variations Summer Term 2014 Lecture 12 ... 1 Numerical solution of the Euler-Lagrange equations ... Calculus of